If $$z$$ is a complex number such that $$|z| \geq 2$$, then the minimum value of $$\left|z + \frac{1}{2}\right|$$:

We are asked to find the least possible value of the modulus $$\left|\,z+\dfrac12\right|$$ when the complex number $$z$$ satisfies the condition $$|z|\ge 2$$.

Write $$z$$ in rectangular form as $$z=x+iy$$, where $$x,y\in\mathbb R$$. The given condition becomes

$$|z|=\sqrt{x^{2}+y^{2}}\;\ge\;2.$$

The quantity whose minimum we seek is

$$\left|\,z+\dfrac12\right|=\left|\,(x+iy)+\dfrac12\right|=\left|\,\bigl(x+\dfrac12\bigr)+iy\right|=\sqrt{\left(x+\dfrac12\right)^{2}+y^{2}}.$$

Instead of working directly with two variables, we can use the Triangle Inequality in its reverse (distance-difference) form. The general statement is:

For any complex numbers $$a$$ and $$b$$, $$\bigl|\,|a|-|b|\,\bigr|\le |a-b|.$$

We apply this formula with $$a=z$$ and $$b=-\dfrac12$$ (note that $$|\, -\dfrac12 \,|=\dfrac12$$):

$$\bigl|\,|z|\;-\;\bigl|-\dfrac12\bigr|\,\bigr|\;\le\;\left|\,z-\Bigl(-\dfrac12\Bigr)\right|=\left|\,z+\dfrac12\right|.$$

Because moduli are non-negative, we may drop the outer absolute value on the left and write

$$|z|-\dfrac12\;\le\;\left|\,z+\dfrac12\right|.$$

Now substitute the information $$|z|\ge 2$$:

$$\left|\,z+\dfrac12\right|\;\ge\;2-\dfrac12=\dfrac32.$$

Thus the expression $$\left|\,z+\dfrac12\right|$$ can never dip below $$\dfrac32$$. The only question left is whether the value $$\dfrac32$$ is actually attainable.

Equality in the reverse Triangle Inequality occurs precisely when the two complex numbers point in the same (or exactly opposite) direction. Concretely, we need $$z$$, the origin, and $$-\dfrac12$$ to lie on the same straight line with $$z$$ farther from the origin than $$-\dfrac12$$. Taking the negative real axis, choose

$$z=-2\quad(\text{so }|z|=2,\;x=-2,\;y=0).$$

Then

$$\left|\,z+\dfrac12\right|=\left|-2+\dfrac12\right|=\left|-\dfrac32\right|=\dfrac32,$$

which meets the lower bound. Hence

$$\min_{\,|z|\ge 2}\;\left|\,z+\dfrac12\right|=\dfrac32.$$

The numerical value $$\dfrac32$$ lies strictly between $$1$$ and $$2$$. Among the offered choices, the statement that the minimum “lies in the interval $$(1,2)$$” is the only one that fits.

Hence, the correct answer is Option 4.